Question

The value of
$3\cdot {}^{15}{C}_2+4\cdot {}^{15}{C}_3+\cdots +16\cdot {}^{15}{C}_{15}=$

• A. $17\times 2^{14}-31$
• B. $17\times 2^{14}$
• C. $15\times 2^{16}-16$
• D. $17\times 2^{14}+16$

Answer 1

The correct option is A $17\times 2^{14}-31$

let's expand the $(1+x{)}^{15}$
$(1+x{)}^{15}={}^{15}{C}_0+{}^{15}{C}_{1}x+{}^{15}{C}_2{x}^2+\cdots +{}^{15}{C}_{15}{x}^{15}$

Multiply by $x$ on both the sides
$\Rightarrow x(1+x{)}^{15}={}^{15}{C}_{0}x+{}^{15}{C}_1{x}^2+{}^{15}{C}_2{x}^3+{}^{15}{C}_3{x}^4+\dots +{}^{15}{C}_{15}{x}^{16}$

Differentiating w.r.t. $x$ on both sides, we get
$\Rightarrow 15x(1+x{)}^{14}+(1+x{)}^{15}={}^{15}{C}_0+2\cdot {}^{15}{C}_{1}x+3\cdot {}^{15}{C}_2{x}^2+4\cdot {}^{15}{C}_3{x}^3+\dots +16\cdot {}^{15}{C}_{15}{x}^{15}$
By putting $x=1,$ we get
$\Rightarrow 15\times 2^{14}+2^{15}={}^{15}{C}_0+2\cdot {}^{15}{C}_1+3\cdot {}^{15}{C}_2+4\cdot {}^{15}{C}_3+\dots +16\cdot {}^{15}{C}_{15}$
$\Rightarrow 2^{14}(15+2)-{}^{15}{C}_0-2\cdot {}^{15}{C}_1=3\cdot {}^{15}{C}_2+4\cdot {}^{15}{C}_3+\dots +16\cdot {}^{15}{C}_{15}$
$\Rightarrow 17\times 2^{14}-31=3\cdot {}^{15}{C}_2+4\cdot {}^{15}{C}_3+\dots +16\cdot {}^{15}{C}_{15}$

Hence, the correct answer is Option b.